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Question Number 27564 by Tinkutara last updated on 09/Jan/18

Commented by Tinkutara last updated on 09/Jan/18

Commented by mrW2 last updated on 12/Jan/18

Commented by mrW2 last updated on 12/Jan/18

(1/2)mv^2 =mgR(1−sin θ) ⇒v^2 =2gR(1−sin θ) mgsin θ−N=((mv^2 )/R)=2mg(1−sin θ) ⇒N=3mg(sin θ−(2/3)) with θ_1 =sin^(−1) (2/3)≈42° for 90°>θ>42°: N>0 for θ<42°: N<0 ⇒Answer 4 is correct.

$$\frac{\mathrm{1}}{\mathrm{2}}{mv}^{\mathrm{2}} ={mgR}\left(\mathrm{1}−\mathrm{sin}\:\theta\right) \\ $$$$\Rightarrow{v}^{\mathrm{2}} =\mathrm{2}{gR}\left(\mathrm{1}−\mathrm{sin}\:\theta\right) \\ $$$${mg}\mathrm{sin}\:\theta−{N}=\frac{{mv}^{\mathrm{2}} }{{R}}=\mathrm{2}{mg}\left(\mathrm{1}−\mathrm{sin}\:\theta\right) \\ $$$$\Rightarrow{N}=\mathrm{3}{mg}\left(\mathrm{sin}\:\theta−\frac{\mathrm{2}}{\mathrm{3}}\right) \\ $$$${with}\:\theta_{\mathrm{1}} =\mathrm{sin}^{−\mathrm{1}} \frac{\mathrm{2}}{\mathrm{3}}\approx\mathrm{42}° \\ $$$${for}\:\mathrm{90}°>\theta>\mathrm{42}°:\:{N}>\mathrm{0} \\ $$$${for}\:\theta<\mathrm{42}°:\:{N}<\mathrm{0} \\ $$$$ \\ $$$$\Rightarrow{Answer}\:\mathrm{4}\:{is}\:{correct}. \\ $$

Commented by Tinkutara last updated on 12/Jan/18

Thank you very much Sir! I got the answer.

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